1 files changed, 46 insertions, 0 deletions
diff --git a/409/CH10/EX10.2/Example10_2.sce b/409/CH10/EX10.2/Example10_2.sce
new file mode 100755
index 000000000..8387a9b46
--- /dev/null
+++ b/409/CH10/EX10.2/Example10_2.sce
@@ -0,0 +1,46 @@
+clear ;
+clc;
+// Example 10.2
+printf('Example 10.2\n\n');
+// Page no. 266
+// Solution
+
+S = 5000 ;// Sulphur [lb]
+// Composition of feed
+CH4 =  80 ;// [%]
+H2S =  20 ;// [%]
+
+n_un = 11 ;// Number of unknowns in the given problem
+n_ie  = 11 ;// Number of independent equations
+d_o_f =  n_un-n_ie ;// Number of degree of freedom
+printf('Number of degree of freedom for the given system is  %i \n',d_o_f);
+m_S = 32.0 ;//molecular wt. of S -[lb]
+mol_S = S/32.0;
+// Extent of reaction can be calculated by using eqn. 9.3 
+// Based on S
+nio_S = 0 ;//[g mol S]
+ni_S = mol_S ;//[g mol S]
+vi_S = 3 ;// coefficint of S -from given reaction
+ex_r = (ni_S-nio_S)/vi_S ;//  Extent of reaction based on S
+printf(' Extent of reaction is %.1f g moles reacting \n',ex_r);
+
+// Product composition 
+vi_H2O = 2 ;// coefficint of H2O
+vi_H2S = -2 ;// coefficint of H2S
+vi_SO2 = -1 ;//coefficint of SO2
+vi_CH4 = 0 ;//coefficint of CH4
+P_H2O = 0+(vi_H2O*ex_r);// [lb mol]
+P_H2S = P_H2O/10 ;//[lb mol]
+P_SO2 = 3*P_H2S ;//[lb mol]
+
+F = (P_H2S-vi_H2S*ex_r)/(H2S/100) ;// total feed-[lb mol]
+F_SO2 = P_SO2-(vi_SO2*ex_r);// feed rate of SO2- [lb mol]
+F_CH4 = (CH4/100)*F+vi_CH4*ex_r ;//feed rate of CH4- [lb mol]
+F_H2S = ((H2S/100)*F) ;// feed rate of H2S-[lb mol]
+
+// We can see from situation that H2S is limiting reagent as ratio of SO2 to H2S in the product gas(3/1) is greater than their molar ratio in chemical reaction(2/1)
+f_cn = -(vi_H2S*ex_r)/((H2S/100)*F) ;// Fractional conversion of limiting reagent
+
+printf('\n(1)Feed rate of H2S-  %.1f lb mol\n',F_H2S);
+printf('(2)Feed rate of SO2-  %.1f lb mol\n',F_SO2);
+printf('(3)Fractional conversion of limiting reagent-  %.2f \n',f_cn);
+\ No newline at end of file